I am trying to find the variance of $\int_t^T(T-s)~dW_s$
I was wondering if this approach is correct:
$$ Var~(\int_t^T(T-s)~dW_s~)=\mathbb E~[~(~\int_t^T(T-s)~dW_s~)^2~]=\mathbb E~[~(\int_0^T(T-s)~1_{s\ge t}dW_s~~)^2~] $$ Then by Ito's Isometry: $$ \int_0^T \mathbb E [(T-s)1_{s\ge t}~]^2~ds $$ Which implies that: $$ =\int_t^T(T-s)^2ds $$
I guess the question boils down to whether or not Ito's Isometry holds for any bounds, not just $ [0,T]$ ?
It is not correct to write
$$\int_0^T (T-s) \, dW_s 1_{s \geq t}$$
since the expression doesn't make sense at all. Instead it should read
$$\int_0^T 1_{s \geq t} \cdot (T-s) \, dW_s.$$
The rest of your calculations is fine.
In fact, Itô's isometry holds for any bounds. Indeed: If $f$ is progressively measurable and $\int_0^T \mathbb{E}(f(s)^2) \, ds < \infty$ for some $T>0$, then $f_{a,b} := f \cdot 1_{[a,b]}$ is still progressively measurable and integrable for $0 \leq a < b \leq T$. Hence, by Itô's isometry,
$$\mathbb{E}\left( \left|\int_a^b f(s) \, dW_s \right|^2 \right) = \mathbb{E} \left( \left| \int_0^T f_{a,b}(s) \, dW_s \right|^2 \right) = \int_0^T \mathbb{E}(f_{a,b}(s)^2) \, ds = \int_a^b \mathbb{E}(f(s)^2) \, ds.$$
Alternatively, we could use the fact that
$$t \mapsto \left(\int_0^t f(s) \, dW_s\right)^2 - \int_0^t f(s)^2 \, ds$$
is a martingale.